Point P lies in the exterior of odot(O, r). o | vedclass

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(C) Given that $P$ is a point outside the circle with center $O$ and radius $r$. $\overleftrightarrow{PQ}$ is a tangent to the circle at point $Q$.By

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$48$

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(C) Given that $P$ is a point outside the circle with center $O$ and radius $r$. $\overleftrightarrow{PQ}$ is a tangent to the circle at point $Q$.By the property of tangents,the radius $OQ$ is perpendicular to the tangent $PQ$ at the point of contact $Q$. Thus,$	riangle OQP$ is a right-angled triangle with $\angle OQP = 90^{\circ}$.In $	riangle OQP$,by the Pythagorean theorem:$OP^2 = OQ^2 + PQ^2$Given $OP = 26$ and $PQ = 10$,we have:$26^2 = r^2 + 10^2$$676 = r^2 + 100$$r^2 = 676 - 100 = 576$$r = \sqrt{576} = 24$.The diameter of the circle is $d = 2r = 2 	imes 24 = 48$.

Point $P$ lies in the exterior of $\odot(O, r)$. $\overleftrightarrow{PQ}$ touches the circle at $Q$. If $PO = 26$ and $PQ = 10$,then find the diameter of the circle.

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